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3 years ago

If x = -6, which inequality is true?

Mathematics

1 answer:

igomit [66]3 years ago

8 0

Answer:

C. 1 - 2x > 13

Step-by-step explanation:

you just substitute

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If a data point has a corresponding z-score of -1.5, then it is one and a half standard deviations above the mean value. 1. True

TEA [102]

Answer: False.

Step-by-step explanation:

For any random variable x,

Formula to calculate the z-score : $z=\dfrac{x-\mu}{\sigma}$

, where $\mu$ = Mean

$\sigma$ = Standard deviation

Let x be the data point has a corresponding z-score of -1.5.

Then , we have

$-1.5=\dfrac{x-\mu}{\sigma}$

$-1.5\sigma=x-\mu$

$\mu-1.5\sigma=x$

i.e. x is one and a half standard deviations below the mean value.

( one and a half = $1+\dfrac{1}{2}=1.5$ )

Therefore , the given statement is false.

8 0

3 years ago

Plz help mee with thisss

timurjin [86]

Answer:

495

Step-by-step explanation:

60x10=600

600-105=495

3 0

3 years ago

Please Help!!!! Solve For L

Sonja [21]

<h3>Given :</h3>

$\tt B = L - L a f$

L = ?

<h3>Solution :</h3>

$\tt B = L - L a f$

By taking L common from $\tt (L - L a f)$ , we get :

$\tt : \implies B = L(1 - a f)$

$\tt : \implies \dfrac{B}{(1 - a f)} = L$

$\tt : \implies L = \dfrac{B}{1 - a f}$

Hence, value of $\tt L = \dfrac{B}{1 - a f}$

4 0

3 years ago

Read 2 more answers

Solve the following initial-value problem, showing all work, including a clear general solution as well as the particular soluti

Vikki [24]

Answer:

General Solution is $y=x^{3}+cx^{2}$ and the particular solution is $y=x^{3}-\frac{1}{2}x^{2}$

Step-by-step explanation:

$x\frac{\mathrm{dy} }{\mathrm{d} x}=x^{3}+3y\\\\Rearranging \\\\x\frac{\mathrm{dy} }{\mathrm{d} x}-3y=x^{3}\\\\\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{3y}{x}=x^{2}$

This is a linear diffrential equation of type

$\frac{\mathrm{d} y}{\mathrm{d} x}+p(x)y=q(x)$ ..................(i)

here $p(x)=\frac{-2}{x}$

$q(x)=x^{2}$

The solution of equation i is given by

$y\times e^{\int p(x)dx}=\int e^{\int p(x)dx}\times q(x)dx$

we have $e^{\int p(x)dx}=e^{\int \frac{-2}{x}dx}\\\\e^{\int \frac{-2}{x}dx}=e^{-2ln(x)}\\\\=e^{ln(x^{-2})}\\\\=\frac{1}{x^{2} } \\\\\because e^{ln(f(x))}=f(x)]\\\\Thus\\\\e^{\int p(x)dx}=\frac{1}{x^{2}}$

Thus the solution becomes

$\tfrac{y}{x^{2}}=\int \frac{1}{x^{2}}\times x^{2}dx\\\\\tfrac{y}{x^{2}}=\int 1dx\\\\\tfrac{y}{x^{2}}=x+c$ $y=x^{3}+cx^{2$

This is the general solution now to find the particular solution we put value of x=2 for which y=6

we have $6=8+4c$

Thus solving for c we get c = -1/2

Thus particular solution becomes

$y=x^{3}-\frac{1}{2}x^{2}$

5 0

3 years ago

Triangle abc is similar to triangle FGH solve for p plebs.

12345 [234]

The value of <em>p </em>is 21

Step-by-step explanation:

ΔABC ~ ΔFGH

$\frac{ab}{fg} = \frac{bc}{gh} \\ \frac{35}{p} = \frac{10}{6} \\ p = \frac{35 \times 6}{10} \\ p = \frac{210}{10} \\ p = 21 \\ \\ or \\ \\ \frac{ab}{fg} = \frac{ac}{fh} \\ \frac{35}{p} = \frac{30}{18} \\ p = \frac{35 \times 18}{30} \\ p = \frac{630}{30} \\ p = 21$

So, the value of <em>p</em> is 21

<em>Hope </em><em>it </em><em>helpful </em><em>and </em><em>useful </em><em>:</em><em>)</em>

5 0

3 years ago